# How to prove $\int_0^{\infty}e^{-sx}\frac{\sin(x)^2}{x}dx = \frac{1}{4}\ln(1- \frac{4}{s^2})$ using Tonelli-Fubini

I need some help with in trying to prove that $$\int_{0}^{\infty}{\rm e}^{-sx}\,\frac{\sin^{2}(x)}{x}\,{\rm d}x = \frac{1}{4}\ln\left(1- \frac{4}{s^{2}}\right) \quad\mbox{using}\ Tonelli\mbox{--}Fubini$$ Now, by Tonelli-Fubini we have that : $$\int_0^{\infty}\int_0^1 e^{-sx}{\sin(2xy)}dydx = \int_0^1\int_0^{\infty}e^{-sx}{\sin(2xy)}dxdy$$ After some calculation, we find that $$\int_0^{\infty}\int_0^1 e^{-sx}{\sin(2xy)}dydx = \int_0^{\infty}e^{-sx}\frac{\sin(x)^2}{x}dx$$ $$\mbox{Now we need to prove that}\ \int_0^1\int_0^{\infty}e^{-sx}{\sin(2xy)}dxdy = \frac{1}{4}\ln\left(1- \frac{4}{s^2}\right)$$

This is when I need help : How can I show this equality? I have been thinking but I cannot come up with anything. I will appreciate some help.

• Do you know the laplace transform of sine? proofwiki.org/wiki/Laplace_Transform_of_Sine
– Zima
Commented 2 days ago
• applying two times integration by parts, you can solve $\int e^{ax}\sin (bx)dx$ Commented 2 days ago
• I think you mean $\frac14\ln(1\color{blue}{+}\tfrac{4}{s^2})$.
– J.G.
Commented 2 days ago
• Use Euler representation of $\sin(bx)$ Commented 2 days ago